What are the solutions to the equation 0=x^3-3x^2-4x+12 ?

The solutions to the equation 0=x^3-3x^2-4x+12 are x=3,x=-2,x=2

Find the zeros of 0=x^3-3x^2-4x+12

The zeros of 0=x^3-3x^2-4x+12 are x=3,x=-2,x=2

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0 = x^{3} - 3 x^{2} - 4 x + 12

x = 3, x = - 2, x = 2

Solution steps

0 = x^{3} - 3 x^{2} - 4 x + 12

Switch sides

x^{3} - 3 x^{2} - 4 x + 12 = 0

Factor

x^{3} - 3 x^{2} - 4 x + 12 : (x - 3) (x + 2) (x - 2)

(x - 3) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x - 3 = 0 : x = 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = 3, x = - 2, x = 2

x, x^{3} + 2 x^{2} = 9 x + 18

- x^{3} + 4 x^{2} + x - 2 = 0

x^{3} + 7 x^{2} + 16 x + 12 = 0

c^{3} = 2

x - 3 x^{5} - \frac{1}{4} = 0

What are the solutions to the equation 0=x^3-3x^2-4x+12 ?
The solutions to the equation 0=x^3-3x^2-4x+12 are x=3,x=-2,x=2
Find the zeros of 0=x^3-3x^2-4x+12
The zeros of 0=x^3-3x^2-4x+12 are x=3,x=-2,x=2